But the title here is totally misleading because it sure sounds like someone took control of 9% of the ipv4 address space but the actual post starts with context.
20 million is a lot, but if you look at geoip, they are around the whole world; I took 3 random latest IPs and I saw Vietnam, Brazil and Angola. So it's not that much when it's worldwide.
But it suggests it's not a geographically limited website. If it's through a website. It's probably not a ad buy. (Who would burn money on that...)
However the requests are literally every second. So it's something very popular. (Or a bot and they are somehow faking the source address...)
Curiously, these are some of the top countries I see when analyzing traffic from malicious scraping bots that disguise themselves as old Chrome versions on my websites.
So it's possible that one of those botnet-ish residential proxy services is being used here. The ones that use things like compromised browser extensions to turn unknowing users into exit nodes.
Edit: Yep, it's residential proxies, someone on the linked page mentioned a website where you can look up the IPs and all of them come up as proxies.
This would make it possible to have thousands of impressions for relatively low amounts of money.
https://github.com/search?q=ipv4.games%2Fclaim&type=code&p=1
While running ads is definitely a possibility, reaching 9% of all available IPs sounds like a crazy expensive campaign. I don't know what the ratio of people to public IP is but I doubt it's one.
The number of times my browser has been hijacked from their ad network is numerous.
Odds are, the culprit owns some IP that is running on 20M devices. Whether it's a mobile game. A bot net. An ad. Or some other script/service that allows other machines to make the request on his/her behalf.
If it’s not meaningful it should be trivial to beat right? ;)
This seems like a super fun game to find the upper bound on IPv4 addresses someone can open a socket from!
For this test to be valid it would need to do much more than just that I think